Question 42: Magnetism and Matter

Question

An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :

Accepted Answer

When an electron is projected perpendicular to a uniform magnetic field $ B $, it travels in a circular path. According to Bohr's quantization rule, the radius of the electron's orbit in the first excited state can be determined using the following approach:
The radius $ r $ in a magnetic field is expressed as:
$    r = \frac{mv}{eB}    $
where $ m $ is the mass of the electron, $ v $ is its velocity, $ e $ is its charge, and $ B $ is the magnetic field strength.
According to Bohr's quantization condition:
$    mvr = \frac{nh}{2\pi}    $
Substituting the expression for $ r $:
$    (eBr)r = \frac{nh}{2\pi}    $
This simplifies to the expression for the radius:
$    r = \sqrt{\frac{nh}{2\pi eB}}    $
In the first excited state, the principal quantum number $ n $ is 2. So, the radius becomes:
$    r = \sqrt{\frac{2h}{2\pi eB}} = \sqrt{\frac{h}{\pi eB}}    $
Thus, for the first excited state, the radius of the electron's orbit is $ \sqrt{\frac{h}{\pi eB}} $.

Answer

$\sqrt{\frac{h}{\pi e B}}$

Answer

$\sqrt{\frac{4 h}{\pi e B}}$

Answer

$\sqrt{\frac{\mathrm{h}}{2 \pi e \mathrm{eB}}}$

Answer

$\sqrt{\frac{2 h}{\pi e B}}$

Magnetism and Matter

An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :

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